# Hypothesis Testing 4/23/2011 What is a Hypothesis?

```4/23/2011
What is a Hypothesis?

Hypothesis Testing
Presented by:
A hypothesis is a claim
population parameter:

population mean
Example: The mean monthly cell phone bill of this city is
μ = \$42

population proportion
Example: The proportion of adults in this city with cell
phones is p = .68
Sources:
Anderson, Sweeney,wiliams, Statistics for Business and Economics , 6e Pearson Education, 2007
Chap 10-2
The Null Hypothesis, H0
The Null Hypothesis, H0
(continued)

States the assumption (numerical) to be
tested

Example: The average number of TV sets in
U.S. Homes is equal to three ( H0 : μ = 3 )

Is always about a population parameter,
H0 : μ = 3



H0 : X = 3
Chap 10-3
The Alternative Hypothesis, H1

Is the opposite of the null hypothesis





Begin with the assumption that the null
hypothesis is true
 Similar to the notion of innocent until
proven guilty
ilt
Refers to the status quo
Always contains “=” , “≤” or “≥” sign
May or may not be rejected
Hypothesis Testing Process
Claim: the
e.g., The average number of TV sets in U.S.
homes is not equal to 3 ( H1: μ ≠ 3 )
population
mean age is 50.
Challenges
g the status q
quo
(Null Hypothesis:
Population
H0: μ = 50 )
Never contains the “=” , “≤” or “≥” sign
May or may not be supported
Is generally the hypothesis that the
researcher is trying to support
Is X= 20 likely if μ = 50?
If not likely,
REJECT
Null Hypothesis
Chap 10-4
Chap 10-5
Now select a
random sample
Suppose
the sample
mean age
is 20: X = 20
Sample
© Mahendra Adhi Nugroho, M.Sc, Accounting Program Study of Yogyakarta State University
For internal use only!
1
4/23/2011
Level of Significance, α
Reason for Rejecting H0
Sampling Distribution of X

Defines the unlikely values of the sample
statistic if the null hypothesis is true

20
If it is unlikely that we
would get a sample
mean of this value ...

X
μ = 50
If H0 is true
... if in fact this were
the population mean…
... then we reject the
null hypothesis that
μ = 50.
Chap 10-7
Is designated by α , (level of significance)

H0: μ = 3
H1: μ ≠ 3
α
α/2
Two-tail test
α/2
Is selected by the researcher at the beginning

Provides the critical value(s) of the test
H0: μ ≤ 3
H1: μ > 3
Rejection
region is
h d d
α

Type I Error
 Reject a true null hypothesis
 Considered a serious type of error
The probability of Type I Error is α
0
Upper-tail test
H0: μ ≥ 3
H1: μ < 3
α
Lower-tail test
Chap 10-8
Errors in Making Decisions
Represents
critical value
0
Typical values are .01, .05, or .10

Level of Significance
and the Rejection Region
Level of significance =
Defines rejection region of the sampling
distribution

Called level of significance of the test

0
Chap 10-9
Errors in Making Decisions
Chap 10-10
Outcomes and Probabilities
(continued)

Possible Hypothesis Test Outcomes
Type II Error
 Fail to reject a false null hypothesis
Actual Situation
Decision
The probability of Type II Error is β
Do Not
Key:
Outcome
(Probability)
Reject
H0
Reject
H0
Chap 10-11
H0 True
H0 False
No error
(1 - α )
Type II Error
(β)
Type I Error
( α)
© Mahendra Adhi Nugroho, M.Sc, Accounting Program Study of Yogyakarta State University
For internal use only!
No Error
(1-β)
Chap 10-12
2
4/23/2011
Factors Affecting Type II Error
Type I & II Error Relationship

 Type
I and Type II errors can not happen at
the same time

Type I error can only occur if H0 is true

Type II error can only occur if H0 is false
If Type I error probability ( α )

, then
Type II error probability ( β )
Chap 10-13



β
when
α

β
when
σ

β
when
n
Chap 10-14
Two Basic Approaches to Hypothesis Testing
The power of a test is the probability of rejecting
a null hypothesis that is false
i.e.,
β
when the difference between
hypothesized parameter and its true value
Power of the Test

All else equal,
¾ There are two basic approaches to conducting a
hypothesis test:
¾ 1- p-Value Approach, and
¾2- Critical Value Approach
Power = P(Reject H0 | H1 is true)
Power of the test increases as the sample size
increases
Slide 16
Chap 10-15
1- p-Value Approach to
One--Tailed Hypothesis Testing
One
2- Critical Value Approach
One--Tailed Hypothesis Testing
One
 Use the Z table to find the critical Z value,
value, and
 Use the equation to find the actual ZZ--Z statistics.
 In order to accept or reject the null hypothesis the p-value is
computed using the test statistic ---Actual
Actual Z value.

 Reject H0 if the
h p-value
l <α
The rejection rule is:
L
Lower
tail:
il Reject
R j
H0 if Actual
A
l z < Critical
C i i l -z α
• Upper tail: Reject H0 if Actual z > Critical zα
•
 Do not reject (accept) H0 if the p-value > α
In other words, if the actual Z (Z statistics) is in the rejection region,
then reject the null hypothesis.
Equation for finding the
actual Z value:
z=
x −μ
σ/ n
Slide 17
© Mahendra Adhi Nugroho, M.Sc, Accounting Program Study of Yogyakarta State University
For internal use only!
Slide 18
3
4/23/2011
Test of Hypothesis
for the Mean (σ Known)
Hypothesis Tests for the Mean

Convert sample result ( x ) to a z value
Hypothesis
Tests for μ
σ Known
Hypothesis
Tests for μ
σ Unknown
σ Known
σ Unknown
Consider the test
The decision rule is:
H0 : μ = μ0
H1 : μ > μ0
x − μ0
> zα
σ
n
Reject H 0 if z =
(Assume the population is normal)
Chap 10-19
Decision Rule
Reject H0 if z =
p-Value Approach to Testing
H0: μ = μ0
x − μ0
> zα
σ
n

H1: μ > μ0
Alternate rule:
α
Reject H0 if X > μ0 + Z ασ/ n
p-value: Probability of obtaining a test
statistic more extreme ( ≤ or ≥ ) than the
p value g
given H0 is true
observed sample


Do not reject H0
Z
0
x
μ0
Chap 10-20
zα
Reject H0
μ0 + z α
Also called observed level of significance
Smallest value of α for which H0 can be
rejected
σ
n
Critical value
Chap 10-21
Chap 10-22
Example: Find Rejection Region
p-Value Approach to Testing
(continued)


Convert sample result (e.g., x ) to test statistic (e.g., z
statistic )
Obtain the p-value
x - μ0
p - value = P(Z >
, given that H0 is true)
 For an upper
σ/ n
t il ttest:
tail
t
= P(Z >

Suppose that α = .10 is chosen for this test
Find the rejection region:
Reject H0
α = .10
x - μ0
| μ = μ0 )
σ/ n
Do not reject H0
Decision rule: compare the p-value to α

If p-value < α , reject H0

If p-value ≥ α , do not reject H0
(continued)

0
1.28
Reject H 0 if z =
Chap 10-23
Reject H0
x − μ0
> 1.28
σ/ n
© Mahendra Adhi Nugroho, M.Sc, Accounting Program Study of Yogyakarta State University
For internal use only!
Chap 10-24
4
4/23/2011
Example: Upper-Tail Z Test
for Mean (σ Known)
Example: Sample Results
(continued)
A phone industry manager thinks that
customer monthly cell phone bill have
increased, and now average over \$52 per
month. The company wishes to test this
claim. (Assume σ = 10 is known)
Obtain sample and compute the test statistic
Suppose a sample is taken with the following
results: n = 64,
64 x = 53.1
53 1 (σ=10 was assumed known)

Form hypothesis test:
H0: μ ≤ 52
the average is not over \$52 per month
H1: μ > 52
the average is greater than \$52 per month
x − μ0
53.1 − 52
=
= 0.88
σ
10
n
64
z=
(i.e., sufficient evidence exists to support the
manager’s claim)
Chap 10-25
Using the sample results,
Example: Decision
Example: p-Value Solution
(continued)
Calculate the p-value and compare to α
Reach a decision and interpret the result:
p-value = .1894
α = .10
0
1.28
z = 0.88
Reject
j
H0
α = .10
Do not reject H0
Do not reject H0 since z = 0.88 < 1.28
i.e.: there is not sufficient evidence that the
mean bill is over \$52
Chap 10-27
H0: μ ≤ 3
= .1894
Chap 10-28
Upper-Tail Tests
In many cases, the alternative hypothesis
focuses on one particular direction
H1: μ > 3
= P(z ≥ 0.88) = 1− .8106
Reject H0
Do not reject H0 since p-value = .1894 > α = .10
One-Tail Tests

1.28
Z = .88
P(x ≥ 53.1| μ = 52.0)
53.1− 52.0 ⎞
⎛
= P⎜ z ≥
⎟
10/ 64 ⎠
⎝
0
Reject H0

There is only one critical
value, since the rejection
area is in only one tail
H0: μ ≤ 3
H1: μ > 3
α
This is an upper-tail
upper tail test since the alternative
hypothesis is focused on the upper tail above the
mean of 3
Do not reject H0
H0: μ ≥ 3
H1: μ < 3
(continued)
(assuming that μ = 52.0)
Reject H0
Do not reject H0
Chap 10-26
This is a lower-tail test since the alternative
hypothesis is focused on the lower tail below the
mean of 3
Z
x
zα
0
Reject H0
μ
Critical value
Chap 10-29
© Mahendra Adhi Nugroho, M.Sc, Accounting Program Study of Yogyakarta State University
For internal use only!
Chap 10-30
5
4/23/2011
Lower-Tail Tests
Two-Tail Tests
H0: μ ≥ 3

In some settings, the
alternative hypothesis does
not specify a unique direction

H1: μ < 3
There is only one critical
value, since the rejection
area is in only one tail
α

Reject H0
Do not reject H0
-zα
Z
0
x
μ
Critical value
H0: μ = 3
H1: μ ≠ 3
α/2
There are two
critical values,
defining the two
regions of
rejection
α/2
x
3
Reject H0
Do not reject H0
-zα/2
Lower
critical value
Chap 10-31
Reject H0
z
+zα/2
0
Upper
critical value
Chap 10-32
Hypothesis Testing Example
Hypothesis Testing Example
(continued)
Test the claim that the true mean # of TV
sets in US homes is equal to 3.
(Assume σ = 0.8)





State the appropriate null and alternative
hypotheses
 H0: μ = 3 , H1: μ ≠ 3
(This is a two tailed test)
Specify the desired level of significance
 Suppose that α = .05 is chosen for this test
Choose a sample size
 Suppose a sample of size n = 100 is selected

Determine the appropriate technique
 σ is known so this is a z test
Set up the critical values
 For α = .05 the critical z values are ±1.96
Collect the data and compute the test statistic
 Suppose the sample results are
n = 100, x = 2.84 (σ = 0.8 is assumed known)
So the test statistic is:
z =
Chap 10-33
X − μ0
σ
n
=
2.84 − 3
− .16
=
= − 2.0
0.8
.08
100
Chap 10-34
Hypothesis Testing Example
Hypothesis Testing Example
(continued)

Is the test statistic in the rejection region?
Reject H0 if z <
-1.96 or z >
1.96; otherwise
do not reject H0
α = .05/2
Reject H0
-z = -1.96
α = .05/2
Do not reject H0
0
(continued)

Reject H0
Reach a decision and interpret the result
α = .05/2
Reject H0
+z = +1.96
-z = -1.96
α = .05/2
Do not reject H0
0
Reject H0
+z = +1.96
-2.0
Since z = -2.0 < -1.96, we reject the null hypothesis and conclude that
there is sufficient evidence that the mean number of TVs in US homes is
not equal to 3
Here, z = -2.0 < -1.96, so the test statistic is
in the rejection region
Chap 10-35
© Mahendra Adhi Nugroho, M.Sc, Accounting Program Study of Yogyakarta State University
For internal use only!
Chap 10-36
6
4/23/2011
t Test of Hypothesis for the Mean
(σ Unknown)
Example: p-Value

Example: How likely is it to see a sample mean of
2.84 (or something further from the mean, in either
direction) if the true mean is μ = 3.0?

Convert sample result ( x ) to a t test statistic
Hypothesis
Tests for μ
x = 2.84 is translated to a z
score off z = -2.0
20
σ Known
α/2 = .025
P(z < −2.0) = .0228
.0228
P(z > 2.0) = .0228
Consider the test
.0228
H0 : μ = μ0
p-value
H1 : μ > μ0
= .0228 + .0228 = .0456
-1.96
-2.0
σ Unknown
α/2 = .025
0
1.96
2.0
Z
Chap 10-37
The decision rule is:
Reject H 0 if t =
(Assume the population is normal)
x − μ0
> t n -1, α
s
n
Chap 10-38
Example: Two-Tail Test
(σ Unknown)
t Test of Hypothesis for the Mean
(σ Unknown)
(continued)

For a two-tailed test:
Consider the test
H0 : μ = μ0
(Assume the population is normal,
and the population variance is
unknown)
k
)
H1 : μ ≠ μ0
The decision rule is:
x − μ0
x − μ0
Reject H 0 if t =
< − t n -1, α/2 or if t =
> t n -1, α/2
s
s
n
n
Chap 10-39
The average cost of a
hotel room in New York
is said to be \$168 per
night.
i ht A random
d
sample
l
of 25 hotels resulted in
x = \$172.50 and
s = \$15.40. Test at the
α = 0.05 level.
(Assume the population distribution is normal)
Example Solution:
Two-Tail Test
H0: μ = 168
H1: μ ≠ 168
 α = 0.05
0 05
Reject H0
-t n-1,α/2
-2.0639
 σ is unknown, so
use a t statistic
 Critical Value:
t24 , .025 = ± 2.0639
α/2=.025
Do not reject H0

Involves categorical variables

Two possible outcomes
Reject H0
0
1.46
x−μ
172.50 − 168
t n −1 =
=
s
15.40
n
25
t n-1,α/2
2.0639

= 1.46
Do not reject H0: not sufficient evidence that
true mean cost is different than \$168
Chap 10-40
Tests of the Population Proportion
α/2=.025
 n = 25
H0: μ = 168
H1: μ ≠ 168
Chap 10-41


“Success”
Success (a certain characteristic is present)

“Failure” (the characteristic is not present)
Fraction or proportion of the population in the
“success” category is denoted by P
Assume sample size is large
© Mahendra Adhi Nugroho, M.Sc, Accounting Program Study of Yogyakarta State University
For internal use only!
Chap 10-42
7
4/23/2011
Proportions
Example: p-Value
(continued)

Sample proportion in the success category is
denoted by p̂


x number of successes in sample
pˆ =
=
n
sample size
Here:

When nP(1 – P) > 9, p̂ can be approximated
by a normal distribution with mean and
standard deviation
P(1− P)

μp̂ = P
σ p̂ =
n
Chap 10-43

If p-value < α , reject H0

If p-value ≥ α , do not reject H0
p-value = .0456
α = .05
α/2 = .025
Since .0456 < .05, we reject
the null hypothesis


H0 False
1.96
2.0
Z
Chap 10-44
Assume the population is normal and the population variance is known.
Consider the test
Do Not
Reject H0
No error
(1 - α )
Type II Error
(β)
Reject H0
Type I Error
( α)
No Error
(1-β)
H0 : μ = μ0
H1 : μ > μ0
Th decision
The
d i i rule
l iis:
Reject H 0 if z =
β denotes the probability of Type II Error
1 – β is defined as the power of the test
x − μ0
> z α or Reject H0 if x = x c > μ0 + Z ασ/ n
σ/ n
If the null hypothesis is false and the true mean is μ*, then the probability of
type II error is
⎛
x −μ* ⎞
⎟
β = P(x < x c | μ = μ*) = P⎜⎜ z < c
σ / n ⎟⎠
⎝
Power = 1 – β = the probability that a false null
hypothesis is rejected
Chap 10-45
Chap 10-46
Type II Error Example

0
Actual Situation
H0 True
Decision
.0228
Type II Error
Recall the possible hypothesis test outcomes:
Key:
Outcome
(Probability)
α/2 = .025
.0228
-1.96
-2.0
Power of the Test

(continued)
Compare the p-value with α
Type II Error Example
Type II error is the probability of failing
to reject a false H0
Suppose we fail to reject H0: μ ≥ 52
when in fact the true&x&c mean is μ
μ* = 50
(continued)

Suppose we do not reject H0: μ ≥ 52 when in fact
the true mean is μ* = 50
This is the range of x where
H0 is not rejected
Thi iis th
This
the ttrue
distribution of x if μ = 50
α
50
52
Reject
H0: μ ≥ 52
Do not reject
H0 : μ ≥ 52
xc
50
Reject
H0: μ ≥ 52
Chap 10-47
52
xc
Do not reject
H0 : μ ≥ 52
© Mahendra Adhi Nugroho, M.Sc, Accounting Program Study of Yogyakarta State University
For internal use only!
Chap 10-48
8
4/23/2011
Type II Error Example
Calculating β
(continued)


Suppose we do not reject H0: μ ≥ 52 when
in fact the true mean is μ* = 50
Here β = P( x ≥
Here,
Suppose n = 64 , σ = 6 , and α = .05
) if μ*
x = 50
(f H0 : μ ≥ 52)
(for
c
So β = P( x ≥ 50.766 ) if μ* = 50
β
α
50
Reject
H0: μ ≥ 52
α
52
xc
σ
6
= 52 − 1.645
= 50.766
n
64
x c = μ0 − z α
50
Do not reject
H0 : μ ≥ 52
50.766
Reject
H0: μ ≥ 52
Chap 10-49
52
Do not reject
H0 : μ ≥ 52
xc
Chap 10-50
Calculating β
Power of the Test Example
(continued)

If the true mean is μ* = 50,
Suppose n = 64 , σ = 6 , and α = .05
⎛
⎞
⎜
50.766 − 50 ⎟
P( x ≥ 50.766 | μ* = 50) = P⎜ z ≥
⎟ = P(z ≥ 1.02) = .5 − .3461 = .1539
6
⎜
⎟
64 ⎠
⎝

The probability of Type II Error = β = 0.1539

The power of the test = 1 – β = 1 – 0.1539 = 0.8461
Actual Situation
Probability of
type II error:
α
β = .1539
50
Reject
H0: μ ≥ 52
Do not reject
H0 : μ ≥ 52
H0 True
Do Not
Reject H0
No error
1 - α = 0.95
Reject H0
52
xc
Key:
Outcome
(Probability)
Decision
Type I Error
α = 0.05
H0 False
Type II Error
β = 0.1539
No Error
1 - β = 0.8461
(The value of β and the power will be different for each μ*)
Chap 10-51
Chap 10-52
Chapter Summary


Performed Z Test for the mean (σ known)

Discussed critical value and p-value approaches to
h
hypothesis
th i ttesting
ti

Performed one-tail and two-tail tests

Performed t test for the mean (σ unknown)

Performed Z test for the proportion

Discussed type II error and power of the test